Past SAT Math, GAT / Qudurat, and AP Calculus AB weekly challenges collected for focused review, solution analysis, and smarter practice.
Each archived challenge was previously published through the StudyGlitch Weekly Math Challenge system. Use the archive to revisit solution paths, compare trap answers, and connect the result to diagnostic-style SAT, GAT, and AP math practice.
41 past weekly challenges found. Open one card at a time for the full solution review.
Solve an SAT advanced algebra question involving common factors and difference of squares.
Advanced Algebra / Factoring by common factor and difference of squaresWhich expression is a factor of y2(x − 3) − 25(x − 3)3?
D.
y + 5x − 15Factor out the common expression (x−3) first.
Start with:
y2(x − 3) − 25(x − 3)3
Factor out (x−3):
(x − 3)[y2 − 25(x − 3)2]
The bracketed expression is a difference of squares:
y2 − [5(x − 3)]2
= [y − 5(x − 3)][y + 5(x − 3)]
One factor is:
y + 5(x − 3) = y + 5x − 15
The expression has a common factor and then a difference-of-squares structure. The listed factor comes from expanding y + 5(x−3).
A common mistake is stopping after factoring out (x−3) and missing the difference of squares.
The trap is not expanding y + 5(x−3) to match the answer choice.
If this was difficult, it may show a weak spot in factoring with grouped expressions.
This SAT advanced algebra question tests factoring structure.
For multi-step factoring, remove the common factor before applying special products.
Recognize 25(x−3)2 as [5(x−3)]2.
Practice an AP Calculus AB definite integral question using parametric equations to change variables.
Definite Integrals / Changing variables in a definite integral using parametric equationsIf x=4cos θ and y=3sin θ, then ∫24 xy dx is equivalent to
D.
48∫0π/3 sin2θ cos θ dθReplace x, y, and dx in terms of θ.
Since x=4cosθ, dx=−4sinθ dθ. Also y=3sinθ, so xy=(4cosθ)(3sinθ)=12sinθcosθ. Thus xy dx = 12sinθcosθ(−4sinθ)dθ = −48sin2θcosθ dθ.
When x=2, 4cosθ=2, so θ=π/3. When x=4, θ=0. Therefore,
∫24xy dx = ∫π/30−48sin2θcosθ dθ = 48∫0π/3sin2θcosθ dθ
The negative from dx is handled by reversing the bounds.
A common mistake is replacing x and y but forgetting to replace dx.
The trap is keeping the original x-bounds after switching to θ.
If this was difficult, it may reveal weakness in changing variables in definite integrals.
This AP Calculus AB definite integral question tests parametric substitution.
For parametric substitutions, convert the bounds and the differential.
The bounds in θ go from π/3 to 0, then reverse them to remove the negative sign.
Practice a hard GAT algebra question simplifying an algebraic fraction using difference of squares.
Algebra / Simplifying algebraic fractions by factoringFor x ≠ 3, simplify:
x2 - 9x - 3
B.
x + 3Factor the numerator using difference of squares.
Factor the numerator:
x2 - 9 = (x - 3)(x + 3)
So:
x2 - 9x - 3 = (x - 3)(x + 3)x - 3
Since x ≠ 3, cancel x - 3:
x + 3
The numerator is a difference of squares, not a simple subtraction. Once it is factored, the common factor x - 3 cancels.
A common mistake is canceling the x terms or subtracting 9 - 3. Only factors can be canceled.
The trap is canceling pieces that are not full factors.
If this was missed, the weak point may be factoring before simplification.
GAT algebra simplification questions often test whether you cancel factors or terms.
In algebraic fractions, factor before canceling.
Recognize x2 - 9 as (x - 3)(x + 3).
Solve an SAT statistics question comparing the median and standard deviation of two dot plots.
Statistics and Data Analysis / Interpreting dot plots and comparing measures of center and spreadThe dot plots represent the distributions of values in data sets A and B. Which of the following statements must be true?
I. The median of data set A is equal to the median of data set B.
II. The standard deviation of data set A is equal to the standard deviation of data set B.
B.
I onlyFirst determine the median of each dot plot, then compare the spread.
Count the values in Data Set A:
4(1), 5(4), 6(2), 7(3), 8(2), 9(4), 10(1)
This is 17 values total, so the median is the 9th value. Because the distribution is centered at 7, the median is 7.
Data Set B has counts:
4(2), 5(4), 6(2), 7(1), 8(2), 9(4), 10(2)
This is also 17 values total, so the median is again the 9th value, which is 7. So Statement I is true.
However, Data Set B has more values farther from the center and fewer values at the center than Data Set A, so Data Set B has a larger standard deviation. Therefore, Statement II is false.
The correct answer is I only.
Both dot plots are symmetric about 7, so they share the same median. But the second distribution places more values at the extremes 4 and 10 and fewer at the center 7, which increases the spread. That means the standard deviations are not equal.
A common mistake is thinking that because both distributions are symmetric and centered at the same value, their standard deviations must also be equal.
The trap is assuming the same center automatically means the same spread.
If this was difficult, it may show a weak spot in comparing measures of center and variability.
This SAT statistics and data analysis question tests median and standard deviation from dot plots.
When comparing dot plots, treat center and spread as separate questions.
For an odd number of data points, the median is the middle position after ordering.
Practice an AP Calculus AB question using the graph of f prime to compare function values.
Applications of Differential Calculus / Using the sign of a derivative graph to compare function valuesFrom the graph it follows that
D.
f(2)<f(3)A function increases where its derivative is positive.
The graph shown is f′. Since f′(x)>0 between x=2 and x=3, the function f is increasing on that interval. Therefore, f(2)<f(3).
The derivative is positive from 0 to 5, so f increases there. The graph does not imply discontinuity of f at 4. Also, f is not decreasing on the entire interval 4<x<7 because f′ is still positive from 4 to 5.
A common mistake is reading the graph as f instead of f′.
The trap is claiming f decreases before f′ becomes negative.
If this was difficult, it may reveal weakness in interpreting derivative graphs.
This AP Calculus AB applications question tests derivative graph interpretation.
Use the sign of f′ to determine where f is increasing or decreasing.
On (2,3), the derivative is positive.
Solve a medium GAT geometry question using similar triangles and proportional side lengths.
Geometry / Using proportional sides in similar trianglesTwo triangles are similar. In the smaller triangle, a side of length 6 corresponds to a side of length 15 in the larger triangle. If another side of the smaller triangle is 10, what is the corresponding side length in the larger triangle?
C.
25Find the scale factor from the smaller triangle to the larger triangle.
The scale factor is:
156 = 52
So the corresponding side is:
10 × 52 = 25
The triangles are similar, so corresponding sides have the same ratio. Since 6 became 15, every smaller-triangle side is multiplied by 52.
A common mistake is adding the difference 15 - 6 = 9 to the other side and getting 19. Similarity uses multiplication, not addition.
The trap is using the difference between sides rather than the scale factor.
If this was missed, the issue may be confusing additive and multiplicative relationships.
GAT similarity questions usually reward ratio thinking instead of additive thinking.
For similar figures, use scale factor, not difference.
Convert 6 → 15 into a multiplier before touching the side 10.
Solve an SAT algebra word problem about a bee colony growth model and remaining population needed.
Algebra / Writing a linear function from a piecewise growth contextA beekeeper's initial observation of the population of a certain bee colony was 1,800 bees. The beekeeper set a goal to increase the population to 3,300 bees. The beekeeper uses a model that predicts the population of this bee colony begins at 1,800 and increases by 120 bees per week in the first two weeks after the initial observation, and then increases by 180 bees per week until the beekeeper's goal is reached. According to this model, at the end of week w after the initial observation, where w > 2, which of the following functions gives the predicted number of bees still needed to reach the beekeeper's goal?
C.
p(w) = 1,620 − 180wFind the population after the first two weeks, then model the remaining amount.
After the first two weeks, the population increases by:
2(120) = 240
So at the end of week 2, the population is:
1,800 + 240 = 2,040
For w > 2, the colony then increases by 180 bees per week. The number of weeks after week 2 is w − 2, so the population is:
2,040 + 180(w − 2)
= 2,040 + 180w − 360 = 1,680 + 180w
The number still needed to reach 3,300 is:
3,300 − (1,680 + 180w) = 1,620 − 180w
The model changes after the first two weeks, so the function for w > 2 must include the progress already made during those first two weeks.
A common mistake is using 3,300 − 180w, which ignores the first two weeks of growth at 120 bees per week.
The trap is applying the 180-bee rate from week 0 instead of after week 2.
If this was difficult, it may show a weak spot in translating piecewise verbal models into formulas.
This SAT algebra question tests linear modeling in a multi-stage context.
For piecewise contexts, account for the earlier interval before writing the later formula.
At week 2, the colony has already gained 240 bees.
Practice an AP Calculus AB limit question about one-sided limits and removable discontinuity.
Limits and Continuity / Using one-sided limits to identify a removable discontinuitySuppose limx→−3− f(x)=−1, limx→−3+ f(x)=−1, and f(−3) is not defined. Which of the following statements is (are) true?
I. limx→−3 f(x)=−1
II. f is continuous everywhere except at x=−3
III. f has a removable discontinuity at x=−3
C.
I and III onlyEqual one-sided limits give a two-sided limit, but they do not describe continuity everywhere else.
Since the left-hand and right-hand limits at x=−3 are both −1, the two-sided limit exists and equals −1. So statement I is true. Since f(−3) is not defined while the limit exists, f has a removable discontinuity at x=−3. So statement III is true. Statement II is not guaranteed because no information is given about continuity at other values of x.
The information is local to x=−3. It determines the limit and the type of discontinuity there, but it cannot prove continuity everywhere else.
A common mistake is assuming statement II is true because the problem only mentions x=−3.
The trap is overgeneralizing from one point.
If this was difficult, it may reveal weakness in local versus global continuity claims.
This AP Calculus AB limits question tests one-sided limits and removable discontinuities.
Do not infer global continuity from local limit information.
Statement II uses the word “everywhere,” which requires more information than given.
Solve a hard GAT algebra parameter question using a given solution value.
Algebra / Finding a parameter from a given solutionThe equation kx - 5 = 3x + 7 has solution x = 4. What is the value of k?
B.
6Substitute x = 4 into the equation, then solve for k.
Substitute x = 4:
4k - 5 = 3(4) + 7
4k - 5 = 12 + 7
4k - 5 = 19
4k = 24
k = 6
The question gives the solution, so use it directly. Substitute 4 for x, then solve the remaining equation for k.
A common mistake is trying to solve for x even though the value of x is already given.
The trap is solving the equation in the wrong direction.
If this felt confusing, the issue may be interpreting parameter questions quickly.
GAT parameter questions test whether you understand what a solution means.
When a problem says “has solution,” substitute that value into the equation.
Use x = 4 immediately.
Solve an SAT percent-change question involving a 179 percent increase followed by a 27 percent decrease.
Statistics and Data Analysis / Determining net percent change after an increase and a decreaseThe value of a painting increased by 179% from the end of 2017 to the end of 2018 and then decreased by 27% from the end of 2019. What was the net percentage increase in the value of the painting from the end of 2017 to the end of 2019?
D.
103.67%Use multipliers for successive percentage changes.
An increase of 179% means the value is multiplied by 2.79. A decrease of 27% means the value is multiplied by 0.73.
2.79 × 0.73 = 2.0367
This means the final value is 203.67% of the original value. Therefore, the net percentage increase is:
203.67% − 100% = 103.67%
Forensic note: The visible wording says “and then decreased by 27% from the end of 2019.” That wording is preserved exactly in the question text. Based on the answer choices and the intended successive-change structure, the calculation is 2.79 × 0.73 = 2.0367, giving a net increase of 103.67%.
A common mistake is subtracting 27% from 179%. Successive percentage changes must be multiplied, not combined by simple subtraction.
The trap is treating the two percentage changes as if they have the same base.
If this was difficult, it may show a weak spot in compound percent-change reasoning.
This is a SAT percent-change question involving two successive changes.
For sequential percent changes, convert each change into a multiplier first.
Use 1 + 1.79 = 2.79 and 1 − 0.27 = 0.73.
Practice an AP Calculus AB continuity question involving removable and nonremovable discontinuities.
Limits and Continuity / Classifying continuity after removable and nonremovable discontinuitiesSuppose
f(x)=3x(x−1)x2−3x+2 for x ≠ 1, 2
f(1)=−3
f(2)=4
Then f(x) is continuous
B.
except at x=2Factor the denominator and simplify where possible.
The denominator factors as x2−3x+2=(x−1)(x−2). For x≠1,2, f(x)=3x(x−1)(x−1)(x−2)=3xx−2. At x=1, the limit is 3(1)1−2=−3, which equals f(1). So f is continuous at x=1. At x=2, the simplified expression has a vertical asymptote, so f is not continuous there. Therefore, f is continuous except at x=2.
The factor x−1 cancels, creating a removable discontinuity that has been filled correctly by f(1)=−3. The factor x−2 remains in the denominator, so x=2 is nonremovable.
A common mistake is saying the function is discontinuous at both 1 and 2 just because the original formula excludes both values.
The trap is not checking the separately defined function values.
If this was difficult, it may reveal weakness in removable versus nonremovable discontinuities.
This AP Calculus AB continuity question tests rational-function discontinuities.
Check whether each excluded value is removable and whether its assigned value fills the hole.
Simplify to 3xx−2 after canceling x−1.
Practice a medium GAT algebra question reading roots from a factored quadratic equation.
Algebra / Reading roots from a factored quadratic expressionIf (x - 4)(x + 7) = 0, what is the sum of the possible values of x?
B.
-3Set each factor equal to zero.
From x - 4 = 0, we get x = 4.
From x + 7 = 0, we get x = -7.
The sum is:
4 + (-7) = -3
The possible values are the roots of the equation. The signs reverse when solving each factor: x - 4 = 0 gives 4, and x + 7 = 0 gives -7.
A common mistake is taking the values as -4 and 7 directly from the parentheses.
The trap is copying signs directly instead of solving each factor.
If this was missed, the issue is likely sign interpretation in factored expressions.
GAT quadratic questions often test whether you read factored form correctly.
In factored form, remember that each factor equals zero, so the signs switch.
Do not expand. The factored form already gives the roots.